Translate the figure 3 units left and 4 units down.

In an examination of purchasing patterns of shoppers, a sample of 16 shoppers revealed that they spent, on average, $54 per hour

Lesechka [4]

Answer:

$54-1.64\frac{21}{\sqrt{16}}=45.39$

$54 +1.64\frac{21}{\sqrt{16}}=62.61$

So on this case the 90% confidence interval would be given by (45.39;62.61)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=54$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=21$ represent the sample standard deviation

n=16 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since the Confidence is 0.90 or 90%, the value of $\alpha=1-0.9=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that $z_{\alpha/2}=1.64$

Now we have everything in order to replace into formula (1):

$54-1.64\frac{21}{\sqrt{16}}=45.39$

$54 +1.64\frac{21}{\sqrt{16}}=62.61$

So on this case the 90% confidence interval would be given by (45.39;62.61)

4 0

3 years ago

Alan sells beaded necklaces. Each large necklace sells for $5.70 and each small necklace sells for $5.10. How much will he earn

inysia [295]

Alan will make $28.50 from 5 large necklaces and $15.30 from 3 small necklaces. $43.80 in total :)

6 0

3 years ago

Read 2 more answers

Joshua Treet had a $400.00 average monthly expenditure for entertainment during the second quarter of the year. He spent $401.50

irga5000 [103]

First, the formula for the average of a data set must be defined. It is calculated by adding all the numbers in the data set and then dividing the sum by the number of data. In this case, the average is set to be equal to $400 with the total number of data being 3, with the September expenditure set as an unknown, x. The equation is then set-up to be: 400 = (401.5 + 250 + x)/3. Thus, Joshua can spend as much as $ 548.5 to be able to have the same average as in his second quarter expenditure.

4 0

3 years ago

Let X denote the distance (m) that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose t

andrezito [222]

Answer and Step-by-step explanation: For an exponential distribution, the probability distribution function is:

f(x) = λ. $e^{-\lambda.x}$

and the cumulative distribution function, which describes the probability distribution of a random variable X, is:

F(x) = 1 - $e^{-\lambda.x}$

(a) Probability of distance at most 100m, with λ = 0.0143:

F(100) = 1 - $e^{-0.0143.100}$

F(100) = 0.76

Probability of distance at most 200:

F(200) = 1 - $e^{-0.0143.200}$

F(200) = 0.94

Probability of distance between 100 and 200:

F(100≤X≤200) = F(200) - F(100)

F(100≤X≤200) = 0.94 - 0.76

F(100≤X≤200) = 0.18

(b) The mean, E(X), of a probability distribution is calculated by:

E(X) = $\frac{1}{\lambda}$

E(X) = $\frac{1}{0.0143}$

E(X) = 69.93

The standard deviation is the square root of variance,V(X), which is calculated by:

σ = $\sqrt{\frac{1}{\lambda^{2}} }$

σ = $\sqrt{\frac{1}{0.0143^{2}} }$

σ = 69.93

Distance exceeds the mean distance by more than 2σ:

P(X > 69.93+2.69.93) = P(X > 209.79)

P(X > 209.79) = 1 - P(X≤209.79)

P(X > 209.79) = 1 - F(209.79)

P(X > 209.79) = 1 - (1 - $e^{-0.0143*209.79}$ )

P(X > 209.79) = 0.0503

(c) Median is a point that divides the value in half. For a probability distribution:

P(X≤m) = 0.5

$\int\limits^m_0 f({x}) \, dx$ = 0.5

$\int\limits^m_0 {\lambda.e^{-\lambda.x}} \, dx$ = 0.5

$\lambda.\frac{e^{-\lambda.x}}{-\lambda}$ = $-e^{-\lambda.x} + e^{0}$

$1 - e^{-\lambda.m}$ = 0.5

$-e^{-\lambda.m}$ = - 0.5

ln( $e^{-0.0143.m}$ ) = ln(0.5)

-0.0143.m = - 0.0693

m = 48.46

6 0

3 years ago

3

Ierofanga [76]

Answer:

A

Step-by-step explanation:

Sum the product of the components in the first row of A with the corresponding components of the first column in B

Repeat this with the components in the second row of A with the corresponding components of the second column in B, that is

AB

= $\left[\begin{array}{ccc}2&1\\3&4\\\end{array}\right]$ $\left[\begin{array}{ccc}3&1\\5&2\\\end{array}\right]$

= $\left[\begin{array}{ccc}2(3)+1(5)&2(1)+1(2)\\3(3)+4(5)&3(1)+4(2)\\\end{array}\right]$

= $\left[\begin{array}{ccc}6+5&2+2\\9+20&3+8\\\end{array}\right]$

= $\left[\begin{array}{ccc}11&4\\29&11\\\end{array}\right]$ → A

6 0

3 years ago